11_11

Materials Science and Engineering: An Introduction

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
254 Alloy Hardness (HRC) Hardness (HRC) 1040 50 <20 5140 55.5 34 4340 57 53 4140 56 47 8620 42 <20 8630 52 28 8640 56 40 8660 64 57 Thus, alloys 4340, 4140, 8640, and 8660 will satisfy the criteria for both surface and center hardnesses. 11.D5 We are asked to determine the maximum diameter possible for a cylindrical piece of 4140 steel that is to be quenched in moderately agitated oil such that the microstructure will consist of at least 50% martensite throughout the entire piece. From Figure 11.5, the equivalent distance from the quenched end of a 4140 steel to give 50% martensite (or a 42.5 HRC hardness) is 26 mm (1-1/16 in.). Thus, the quenching rate at the center of the specimen should correspond to this equivalent distance. Using Figure 11.8(b), the center specimen curve takes on a value of 26 mm (1-1/16 in.) equivalent distance at a diameter of about 75 mm (3 in.). 11.D6 In this problem we are asked to describe a heat treatment that may be used on a 45 mm diameter steel shaft of a 1040 steel such that it will have a uniform tensile strength of at
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: least 620 MPa across the entirety of its cross-section. First of all, if the steel is heat treated so as to produce martensite or tempered martensite, there will undoubtedly be a variation of tensile strength over the cross-section. A better heat treatment would be an isothermal one. From Equation (6.20a) a tensile strength of 620 MPa corresponds to a Brinell hardness of about 180. Upon consultation of Figure 10.22(a), we note that for an alloy of composition of 0.40 wt% C, in order to achieve a hardness of 180 BHN, a microstructure of fine pearlite is required. In Figure 10.29 is shown an isothermal transformation diagram for a 0.45 wt% C alloy, which would be very similar to that for a 1040 steel. According to this diagram, we would want to austenitize the alloy at approximately 800∞C, rapidly cool to about 600∞C, and hold the alloy at this temperature for on the order of at least 3 s, after which it is cooled to room temperature....
View Full Document

This document was uploaded on 05/04/2010.

Ask a homework question - tutors are online